mark scheme 4721 core mathematics 1 june 2014 … mark scheme is published as an aid to teachers and...

Oxford Cambridge and RSA Examinations

GCE

Mathematics

Unit 4721: Core Mathematics 1

Advanced Subsidiary GCE

Mark Scheme for June 2014

OCR (Oxford Cambridge and RSA) is a leading UK awarding body, providing a wide range of qualifications to meet the needs of candidates of all ages and abilities. OCR qualifications include AS/A Levels, Diplomas, GCSEs, Cambridge Nationals, Cambridge Technicals , Functional Skills, Key Skills, Entry Level qualifications, NVQs and vocational qualifications in areas such as IT, business, languages, teaching/training, administration and secretarial skills. It is also responsible for developing new specifications to meet national requirements and the needs of students and teachers. OCR is a not-for-profit organisation; any surplus made is invested back into the establishment to help towards the development of qualifications and support, which keep pace with the changing needs of today’s society. This mark scheme is published as an aid to teachers and students, to indicate the requirements of the examination. It shows the basis on which marks were awarded by examiners. It does not indicate the details of the discussions which took place at an examiners’ meeting before marking commenced. All examiners are instructed that alternative correct answers and unexpected approaches in candidates’ scripts must be given marks that fairly reflect the relevant knowledge and skills demonstrated. Mark schemes should be read in conjunction with the published question papers and the report on the examination. OCR will not enter into any discussion or correspondence in connection with this mark scheme. © OCR 2014

4721 Mark Scheme June 2014

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1. Annotations and abbreviations

Annotation in scoris Meaning

Blank Page – this annotation must be used on all blank pages within an answer booklet (structured or unstructured) and on each page of an additional object where there is no candidate response.

and BOD Benefit of doubt

FT Follow through ISW Ignore subsequent working

M0, M1 Method mark awarded 0, 1 A0, A1 Accuracy mark awarded 0, 1

B0, B1 Independent mark awarded 0, 1 SC Special case

Omission sign MR Misread

Highlighting Other abbreviations in mark scheme

Meaning

E1 Mark for explaining

U1 Mark for correct units G1 Mark for a correct feature on a graph

M1 dep* Method mark dependent on a previous mark, indicated by *

cao Correct answer only oe Or equivalent

rot Rounded or truncated soi Seen or implied

www Without wrong working


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2. Subject-specific Marking Instructions for GCE Mathematics Pure strand

a Annotations should be used whenever appropriate during your marking. The A, M and B annotations must be used on your standardisation scripts for responses that are not awarded either 0 or full marks. It is vital that you annotate standardisation scripts fully to show how the marks have been awarded.

For subsequent marking you must make it clear how you have arrived at the mark you have awarded

b An element of professional judgement is required in the marking of any written paper. Remember that the mark scheme is designed to assist in marking incorrect solutions. Correct solutions leading to correct answers are awarded full marks but work must not be judged on the answer alone, and answers that are given in the question, especially, must be validly obtained; key steps in the working must always be looked at and anything unfamiliar must be investigated thoroughly. Correct but unfamiliar or unexpected methods are often signalled by a correct result following an apparently incorrect method. Such work must be carefully assessed. When a candidate adopts a method which does not correspond to the mark scheme, award marks according to the spirit of the basic scheme; if you are in any doubt whatsoever (especially if several marks or candidates are involved) you should contact your Team Leader.

c The following types of marks are available. M

A suitable method has been selected and applied in a manner which shows that the method is essentially understood. Method marks are not usually lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, eg by substituting the relevant quantities into the formula. In some cases the nature of the errors allowed for the award of an M mark may be specified. A

Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated Method mark is earned (or implied). Therefore M0 A1 cannot ever be awarded. B

Mark for a correct result or statement independent of Method marks.


3

E

A given result is to be established or a result has to be explained. This usually requires more working or explanation than the establishment of an unknown result. Unless otherwise indicated, marks once gained cannot subsequently be lost, eg wrong working following a correct form of answer is ignored. Sometimes this is reinforced in the mark scheme by the abbreviation isw. However, this would not apply to a case where a candidate passes through the correct answer as part of a wrong argument.

d When a part of a question has two or more ‘method’ steps, the M marks are in principle independent unless the scheme specifically says otherwise; and similarly where there are several B marks allocated. (The notation ‘dep *’ is used to indicate that a particular mark is dependent on an earlier, asterisked, mark in the scheme.) Of course, in practice it may happen that when a candidate has once gone wrong in a part of a question, the work from there on is worthless so that no more marks can sensibly be given. On the other hand, when two or more steps are successfully run together by the candidate, the earlier marks are implied and full credit must be given.

e The abbreviation ft implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A and B marks are given for correct work only — differences in notation are of course permitted. A (accuracy) marks are not given for answers obtained from incorrect working. When A or B marks are awarded for work at an intermediate stage of a solution, there may be various alternatives that are equally acceptable. In such cases, exactly what is acceptable will be detailed in the mark scheme rationale. If this is not the case please consult your Team Leader. Sometimes the answer to one part of a question is used in a later part of the same question. In this case, A marks will often be ‘follow through’. In such cases you must ensure that you refer back to the answer of the previous part question even if this is not shown within the image zone. You may find it easier to mark follow through questions candidate-by-candidate rather than question-by-question.

f Wrong or missing units in an answer should not lead to the loss of a mark unless the scheme specifically indicates otherwise. Candidates are expected to give numerical answers to an appropriate degree of accuracy, with 3 significant figures often being the norm. Small variations in the degree of accuracy to which an answer is given (e.g. 2 or 4 significant figures where 3 is expected) should not normally be penalised, while answers which are grossly over- or under-specified should normally result in the loss of a mark. The situation regarding any particular cases where the accuracy of the answer may be a marking issue should be detailed in the mark scheme rationale. If in doubt, contact your Team Leader.


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g Rules for replaced work

If a candidate attempts a question more than once, and indicates which attempt he/she wishes to be marked, then examiners should do as the candidate requests.

If there are two or more attempts at a question which have not been crossed out, examiners should mark what appears to be the last (complete) attempt and ignore the others. NB Follow these maths-specific instructions rather than those in the assessor handbook.

h For a genuine misreading (of numbers or symbols) which is such that the object and the difficulty of the question remain unaltered, mark according to the scheme but following through from the candidate’s data. A penalty is then applied; 1 mark is generally appropriate, though this may differ for some units. This is achieved by withholding one A mark in the question. Note that a miscopy of the candidate’s own working is not a misread but an accuracy error.


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Question Answer Marks Guidance

1 5x2 + 10x + 2 = 5(x

2 + 2x) + 2

= 5[(x + 1) 2– 1] + 2

= 5(x + 1) 2

– 3

B1

B1

M1

A1

[4]

p = 5 q = 1

2 –5 “their q”2 or

2

5– “their q”

2 Must be

evidence of squaring r = –3

If p, q and r found correctly, then ISW slips in format. 5(x + 1)

2 + 3 B1 B1 M0 A0

5(x + 1) – 3 B1 B1 M1 A1 (BOD) 5(x + 1 x)

2 – 3 B1 B0 M1 A0

5(x2 + 1)

2 – 3 B1 B0 M1 A0

5(x – 1)2 – 3 B1 B0 M1 A0

5 x (x + 1)2 – 3 B0 B1M1A0

2 i) 2 3 B1 [1]

cao

Do not accept 6 3

3

ii) 10 3 18 3

8 3

B1

B1

[2]

27 3 3 soi , not just 9 3

iii) 5 122 23 3 3

9 3

B1

B1

[2]

Separate 3 from

5

23 Allow only

1

23 3 3 , 23 3 ,3 3 3 , or

81 3 , 3 9 3 for first mark

3 2

24 3 1 0

(4 1)( 1) 0

1, 1

4

1

4

1

2

k x

k k

k k

k k

x

x

M1*

M1dep*

A1

M1

A1

[5]

Substitute for 2x

Attempt to solve resulting quadratic

Correct values of k soi

Attempt to square root

Final answers correct, no extras

No marks if whole equation square rooted etc. No marks if straight to formula with no evidence of substitution at start and no square rooting/squaring at end. If factorising into two brackets: (4x

2 – 1)(x

2 + 1) = 0 M1 A1

(2x + 1) (2x – 1)(x2 + 1) = 0 M1 A1

A1 as before

Spotted solutions: If M0 DM0 or M1 DM0

SR B1 x = 1

2www

SC B1 x = 1

2 www

(Can then get 5/5 if both found www and


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exactly two solutions justified)

4 i) (2, 7) B1

[1]

ii) (1, 5) B1

[1]

4 iii) Translation

– 4 units parallel to the x axis B1 B1

[2]

Translation Correct description e.g. correct vector (not as a coordinate), “4 units to the left”

Do not allow second B1 after incorrect

type of transformation e.g.

stretch/rotation etc. but allow after

shift/move etc.

Do not accept shift/move etc. for first B1 For “parallel to the x axis” allow “horizontally”, “in the x direction”. Do not accept “in/on/ across/up/along/to/towards the x axis”. Do not accept “factor 4” etc. Allow extra if not incorrect.

5 i) 5–3< 6x < 14–3 2< 6x < 11

1

3< x <

11

6

M1

A1

A1

[3]

Attempt to solve two equations/inequalities each involving all

3 terms 2, 11 seen from correct inequalities

www Award full marks if initially working with equations but final answer correct.

Allow “ 1

3< x and x <

11

6” “

1

3< x , x

< 11

6” but do not allow “

1

3< x or x <

11

6”

ii) 23 13 10 0x x

(3 2)( 5) 0x x

2, 5

3x x

M1*

M1dep*

A1

M1

A1 [5]

Expands and rearranges to collect all terms on one side Correct method to find roots

2,5

3 seen as roots

Chooses “outside region” for their roots of their quadratic Do not allow strict inequalities for final mark

See guidance at end of mark scheme

e.g. 2

53

x scores M1A0

Allow “2

, 53

x x ”, “2

3x

or 5x ” but do not allow “2

3x


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and 5x ” SC If question “misread” as

(3 13) 0x x

Roots found as 0, 13

3B1

130,

3x x etc. as above B1, max

2/5

6 i) 13 2

32 2

6 4 5

18 2 5

y x x x

dyx x

dx

B1

M1

A1 A1

[4]

1

24

4xx

soi

Attempt to differentiate, any term correct Two correct terms Fully correct, no “+c”

6 ii) 52

22

36 3d y

x xdx

M1

A1

[2]

Attempt to differentiate their dy

dx

cao www in either part

Any term still involving x correct – follow through from their expression for the M mark only

7 i) 5 1 7 5,

2 2

(2, 1)

M1

A1

[2]

Correct method to find midpoint of line

At least 3 out of 4 terms correctly substituted

ii) Gradient of AB =

7 52

5 1

Perpendicular gradient =1

2

17 ( 5)

2

2 19 0

y x

x y

B1

B1ft

M1

A1ft

A1

[5]

Gradient of AB correctly found as 2

Fully processed 1

their gradient

Equation of straight line through A or B, any non-zero gradient Equation of straight line through A only, their perpendicular gradient, in any form Correct equation in given form

i.e. ( 2 19) 0k x y for integer k .

Must have “=0”.


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8 i) 2 29 7 2

dyx x

dx

When x = 1, 9 7 2 0dy

dx

Therefore a stationary point

M1*

A1

A1

M1dep

A1

[5]

Attempt to differentiate, any term correct Two correct terms Fully correct Substitute x = 1 into their derivative Correctly obtain zero www and state conclusion AG

Alternative for the last two marks: Sets derivative to zero and makes valid attempt to solve resulting quartic M1dep

Correctly establishes x = 1 as solution and draws clear conclusion A1www

8 ii) 23

218 4

d yx x

dx

When x =1,

2

2

d y

dx>0 so minimum

M1

A1

[2]

Correct method to find nature of stationary point e.g. substituting x = 1 into second derivative (at least one term correct from their first derivative in (i) ) No incorrect working seen in this part i.e. if second derivate is evaluated, it must be 22.

Alternate valid methods include: 1) Evaluating gradient at either side of 1(x >0) 2) Evaluating y at 1 and either side of 1 (x >0) If using alternatives, working must be fully correct to obtain the A mark

8 iii) When x=1, y = –2 (0, –2)

B1

B1

[2]

Finding y = –2 at x = 1 Correct coordinate www

9 i) y coordinate of the centre is –5 Radius = 5

B1

B1

Correct y value Correct radius

Alt Shows only meets x axis at one point B1

Centre is five units below x axis and radius is five, so just touches the x-axis

B1

[3]

Correct explanation based on the above – allow clear diagram www

Correct y value for the centre B1 Correct explanation B1 www

9 ii) 2 2 2

2 2

2

(6 2) ( 5)

25 16 10 25 25

10 16 0

CP k

CP k k

k k

(k + 2)(k + 8)<0 –8< k < –2

M1

A1

A1

M1

A1 [5]

Attempt to find CP or CP2

Correct three term quadratic expression* k = –2 and k = –8 found Chooses “inside region” for their roots of their quadratic Must be strict inequalities for the A mark * Or (k + 5)

2 < 9

Alternative Puts x = 6 to into equation of circle M1

Correct three term quadratic equation*, could be in terms of y A1

k = –2 and k = –8 found (allow y) A1 Then as main scheme * Or (k + 5)

2 = 9

SC Trial and improvement B2 if final answer correct (B1 if inequalities are not strict)


9


Can only get 5/5 if fully explained

iii) (2y – 2)2 + (y + 5)

2 = 25

5y2 + 2y + 4=0

b2–4ac = 4–454

= –76 < 0 ,so line and circle do not meet

M1*

A1

M1dep*

A1

[4]

Attempts to eliminate x or y from equation of circle Correct three term quadratic obtained Correct method to establish quadratic has no roots e.g. considers value of b

2–4ac,

tries to find roots from quadratic formula Correct clear conclusion www AG

If y eliminated: 5x

2 + 4x + 16=0

b2–4ac = 16–4516

= –304 No marks for purely graphical

attempts

10

i)

B1

B1

B1

[3]

Positive cubic with max and min Correct y intercept – graph must be drawn Double root shown at x = – 2 and single

root at x = 2

3with no extras – graph must

be drawn

For first mark must clearly be a cubic – must not stop at either axis, do not allow straight line sections/tending to extra turning points etc.

ii) x2+ 4x + 4 or 2x

2+ x – 6

2x

3+ 5x

2 – 4 x – 12

26 10 4dy

x xdx

When x = – 1, gradient = –8 When x = – 1, y = –5 y + 5= –8(x + 1) 8x + y + 13 = 0

B1

M1

A1

M1*

M1dep*

A1ft

B1

M1

A1

Obtain one quadratic factor Multiply their three term quadratic by linear factor to obtain at least 5 term cubic If simplified, must be correct Attempt to differentiate (power of at least one term involving x reduced by one) Substitutes to find gradient at x = –1 Correct gradient found ft their derivative, differentiation of their expression must be fully correct to earn this mark Correct y value Correct equation of straight line through (–1, their y), their gradient from differentiation

Check for working for this in 10 (i) Alternative using product rule: Clear attempt at product rule M1* Differentiates (x + 2)

2 correctly A1

Both expressions fully correct A2 (1

each), then as main scheme

y must have been found, do not allow use of gradient of normal instead of tangent

i.e. (8 13) 0k x y . Must have

“=0”.


10


[9] Correct answer in correct form Note If x = 1 used instead of x = –1 , then max possible from last 5 marks is M1

M1 only


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APPENDIX 1

Solving a quadratic This is particularly important to mark correctly as it features several times on the paper. Consider the equation: 3x

2 – 13x – 10 = 0

1) If the candidate attempts to solve by factorisation, their attempt when expanded must produce the correct quadratic term and one other correct term (with correct sign): (3x + 5)(x – 2) M1 3x

2 and – 10 obtained from expansion

(3x – 4)(x – 3) M1 3x2 and – 13x obtained from expansion

(3x + 5)(x + 2) M0 only 3x2 term correct

2) If the candidate attempts to solve by using the formula

a) If the formula is quoted incorrectly then M0. b) If the formula is quoted correctly then one sign slip is permitted. Substituting the wrong numerical value for a or b or c scores M0

213 ( 13) 4 3 10

2 3

earns M1 (minus sign incorrect at start of formula)

213 ( 13) 4 3 10

2 3

earns M1 (10 for c instead of −10 is the only sign

slip)

213 ( 13) 4 3 10

2 3

M0 (2 sign errors: initial sign and c incorrect)

213 ( 13) 4 3 10

2 10

M0 (2c on the denominator instead of 2a)

Notes – for equations such as 3x

2 – 13x – 10 = 0, then b

2 = 13

2 would be condoned in the discriminant and

would not be counted as a sign error. Repeating the sign error for a in both occurrences in the formula would be two sign errors and score M0.

c) If the formula is not quoted at all, substitution must be completely correct to earn the M1

3) If the candidate attempts to complete the square, they must get to the “square root stage” involving ±; we are looking for evidence that the candidate knows a quadratic has two solutions!

2

2

2

2

3 13 10 0

133 10 0

3

13 1693 10 0

6 36

13 289

6 36

13 289

6 36

x x

x x

x

x

x

This is where the M1 is awarded – arithmetical errors may be condoned

provided 13

6x seen or implied


12

If a candidate makes repeated attempts (e.g. fails to factorise and then tries the formula), mark only what you consider to be their last full attempt.

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mark scheme 4721 core mathematics 1 june 2014 … mark scheme is published as an aid to teachers and...

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